Boost converter inductor current equation

The Resistive load is considered. The working of a buck-boost regulator is explained using the circuit diagram as shown in Figure 1. The circuit diagram for Mode of Operation Interval 1 is shown in Figure 3 and the corresponding waveforms are shown in Figure 2. From the waveform of voltage across the inductor, as shown in Figure 2, the equation for the inductor voltage write as. When a load is reapplied the inductor needs to recharge, and so the transistor's duty cycle increases pulling the inductor towards ground, and because of the increased duty cycle V out decreases when we really want it to increase.

This causes an instability, which is well known for boost converters, and not a problem with buck converters. One way to combat this instability is to choose a large enough inductor so that the ripple current is greater than twice the minimum load current.

When this condition is met then the inductor is always in continuous mode. These design equations have been incorporated into a convenient Switching Converter Calculator. All rights Reserved. And no, it does not involve the conversion of DC to AC and back again. As it involves too many steps.

Anything that has too many steps is inefficient; this is a good life lesson too. A boost converter is one of the simplest types of switch mode converter. As the name suggests, it takes an input voltage and boosts or increases it. Also needed is a source of a periodic square wave. As you can see, there are only a few parts required to make a boost converter. It is less cumbersome than an AC transformer or inductor.

It was a requirement that these converters be as compact and as efficient as possible. To understand the working of a boost converter, it is mandatory that you know how inductors , MOSFETs, diodes and capacitors work. With that knowledge, we can go through the working of the boost converter step by step.

Here, nothing happens. The output capacitor is charged to the input voltage minus one diode drop. Also, a magnetic field builds up around the inductor. Note the polarity of the voltage applied across the inductor.

So it does not like the sudden turning off of the current. It responds to this by generating a large voltage with the opposite polarity of the voltage originally supplied to it using the energy stored in the magnetic field to maintain that current flow. If we forget the rest of the circuit elements and notice only the polarity symbols, we notice that the inductor now acts like a voltage source in series with the supply voltage. This means that the anode of the diode is now at a higher voltage than the cathode remember, the cap was already charged to supply voltage in the beginning and is forward biased.

The output capacitor is now charged to a higher voltage than before, which means that we have successfully stepped up a low DC voltage to a higher one! I recommend that you go through the steps once again very slowly and understand them intuitively. Podcast Helping communities build their own LTE networks.

Podcast Making Agile work for data science. Featured on Meta. New post summary designs on greatest hits now, everywhere else eventually. Related Hot Network Questions. Question feed. Accept all cookies Customize settings.